#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll NEG = LLONG_MIN / 4;
struct Line {
ll m; // slope
ll b; // intercept
int id; // index that produced this line
Line(ll _m=0, ll _b=0, int _id=0): m(_m), b(_b), id(_id) {}
};
bool bad(const Line &l1, const Line &l2, const Line &l3){
// check if l2 is unnecessary between l1 and l3
// (b3 - b1) / (m1 - m3) <= (b2 - b1) / (m1 - m2)
// cross multiply, use __int128 to avoid overflow
__int128 left = (__int128)(l3.b - l1.b) * (__int128)(l1.m - l2.m);
__int128 right = (__int128)(l2.b - l1.b) * (__int128)(l1.m - l3.m);
return left <= right;
}
struct CHT {
vector<Line> hull;
int ptr = 0; // pointer for queries with increasing x
void clear(){ hull.clear(); ptr = 0; }
void add(Line L){ // slopes must be non-decreasing
while(hull.size() >= 2 && bad(hull[hull.size()-2], hull.back(), L))
hull.pop_back();
hull.push_back(L);
if (ptr >= (int)hull.size()) ptr = hull.size() - 1;
}
// query for maximum at x; x queries must be non-decreasing across calls
pair<ll,int> query(ll x){
if (hull.empty()) return {NEG, 0};
while(ptr + 1 < (int)hull.size()){
__int128 v1 = (__int128)hull[ptr].m * x + (__int128)hull[ptr].b;
__int128 v2 = (__int128)hull[ptr+1].m * x + (__int128)hull[ptr+1].b;
if (v2 >= v1) ++ptr;
else break;
}
ll val = (ll)((__int128)hull[ptr].m * x + (__int128)hull[ptr].b);
return {val, hull[ptr].id};
}
};
// build and return CHT hull for level (K-1) using items i = 1..upto
// (we follow the same insertion rule as original: when computing dp_j at layer j, we insert into lc[j] only if j < K)
CHT build_lc_last(const vector<ll> &pre, int n, int K, int upto){
// K is the current k we want to query for (so we want lc[K-1] with lines from dp_{K-1})
vector<CHT> lc(K+1); // indices 0..K
lc[0].add(Line(0, 0, 0)); // base
for (int i = 1; i <= upto; ++i){
int mxj = min(i, K);
// iterate j desc to avoid using newly inserted lines in same i
for (int j = mxj; j >= 1; --j){
auto g = lc[j-1].query(pre[i] - pre[n]);
if (g.first == NEG) continue;
ll dp = g.first + pre[i] * (pre[n] - pre[i]);
if (j < K) lc[j].add(Line(pre[i], dp, i));
}
}
// return copy of lc[K-1]
return lc[K-1];
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, k;
if (!(cin >> n >> k)) return 0;
vector<ll> a(n+1);
vector<ll> pre(n+1, 0);
for (int i = 1; i <= n; ++i){
cin >> a[i];
pre[i] = pre[i-1] + a[i];
}
ll res = NEG;
int bestPos = 0;
// initial full run to find best result and bestPos
vector<CHT> lc(k+1);
lc[0].add(Line(0, 0, 0));
for (int i = 1; i <= n; ++i){
int mxj = min(i, k);
for (int j = mxj; j >= 1; --j){
auto g = lc[j-1].query(pre[i] - pre[n]);
if (g.first == NEG) continue;
ll dp = g.first + pre[i] * (pre[n] - pre[i]);
if (j < k) lc[j].add(Line(pre[i], dp, i));
if (j == k && i < n && dp > res){
res = dp;
bestPos = i;
}
}
}
if (res == NEG) {
// If there's no valid splitting (edge cases), print something reasonable:
// Depending on problem, could be 0 and no cuts; here print 0 and nothing.
cout << 0 << "\n";
return 0;
}
cout << res << "\n";
// reconstruct cuts by recomputing hulls up to bestPos-1 for each step
vector<int> cuts;
int rem = k;
int curPos = bestPos;
while (rem > 0 && curPos > 0){
cuts.push_back(curPos);
// build hulls up to level rem-1 using items up to curPos-1
if (rem == 0) break;
CHT ch = build_lc_last(pre, n, rem, curPos - 1);
// query ch at x = pre[curPos] - pre[n] to get predecessor id
auto g = ch.query(pre[curPos] - pre[n]);
int prev = g.second; // might be 0 (base)
curPos = prev;
--rem;
}
reverse(cuts.begin(), cuts.end());
for (int i = 0; i < (int)cuts.size(); ++i){
if (i) cout << ' ';
cout << cuts[i];
}
cout << '\n';
return 0;
}
